Maximum Efficiency of a Solar-energy Panel

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SUMMARY

The maximum efficiency of solar-energy panels currently stands at 12%. In the southwestern United States, the average solar intensity is approximately 1.0 kW/m². To meet the energy requirements of the United States, estimated at 5.00 x 10²⁰ J/y, a significant area of solar panels must be deployed. The calculations involve using the equations for energy conversion and area estimation based on solar intensity and efficiency.

PREREQUISITES
  • Understanding of solar energy conversion efficiency
  • Familiarity with energy units, specifically Joules and Watts
  • Basic knowledge of area calculations in square meters
  • Proficiency in using equations for energy requirements and solar intensity
NEXT STEPS
  • Research the latest advancements in solar panel efficiency beyond 12%
  • Learn about solar energy conversion calculations using Joules and Watts
  • Explore the geographical impact on solar intensity in different regions
  • Investigate the area requirements for solar farms to meet large-scale energy demands
USEFUL FOR

Engineers, renewable energy researchers, and policymakers focused on solar energy solutions and infrastructure development.

alweltz
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Homework Statement



The maximum efficiency of a solar-energy panel in converting solar energy into useful electrical energy is currently about 12 percent. In a region such as the southwestern United States the solar intensity reaching Earth's surface is about 1.0 kW/m2 on average during the day. Estimate the area that would have to be covered by solar panels in order to supply the energy requirements of the United States (approximately 5.00 1020 J/y).


Homework Equations



(5.00 x 10^20 J/y) X (1/365 y/d) [eq a.]

A= (L) x (w) [eq b.]

(1.0 kW/m^2) x (.12) [eq c.]

The Attempt at a Solution



I have tried multiplying [eq c.] by [eq. a]. Then, take that answer and multiply it by itself to get m^2.
 
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alweltz said:

Homework Statement



The maximum efficiency of a solar-energy panel in converting solar energy into useful electrical energy is currently about 12 percent. In a region such as the southwestern United States the solar intensity reaching Earth's surface is about 1.0 kW/m2 on average during the day. Estimate the area that would have to be covered by solar panels in order to supply the energy requirements of the United States (approximately 5.00 1020 J/y).


Homework Equations



(5.00 x 10^20 J/y) X (1/365 y/d) [eq a.]

A= (L) x (w) [eq b.]

(1.0 kW/m^2) x (.12) [eq c.]

The Attempt at a Solution



I have tried multiplying [eq c.] by [eq. a]. Then, take that answer and multiply it by itself to get m^2.

Do you know how to convert Joules per unit time to watts?
 

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